Search in Rotated Sorted Array:在有转折的升序数列中搜索

Suppose an array sorted in ascending order is rotated at some pivot unknown to you beforehand.
(i.e., 

0 1 2 4 5 6 7

 might become 

4
5 6 7 0 1 2

).
You are given a target value to search. If found in the array return its index, otherwise return -1.
You may assume no duplicate exists in the array.

思路：二分搜索，只不过因为有转折，所以判断那一部分是正常的顺序，那一部分有转折。如果target在正常部分，就正常搜索。

注意边界的判断，比如[1,3] target = 3,[1,2],target = 0，[1] target = 1这种。

class Solution {
public int search(int[] nums, int target) {
if(nums.length==0) return -1;
int s = 0;
int e = nums.length-1;
int m;
while(s<=e){
m = (s+e)/2;
System.out.println("nums[m]="+nums[m]+";m="+m);
if(nums[m]==target) {
return m;
}
//System.out.println("m="+m+";s="+s+";e="+e);
if(nums[s]<=nums[m]){//s到m之间正常有序，如果target在此区间内，正常二分,等号的目的是考虑[3,1]target=1的情况
if(nums[m]>target&&target>=nums[s]){
e = m-1;
}else{
s = m+1;
}
}else{//m到e之间正常有序，如果target在此区间内，正常二分
if(nums[e]>=target&&target>nums[m]){
s = m+1;
}else{
e = m-1;
}

}
}

return -1;
}
}注意，这几个边界条件我是硬试出来的

。。。晕了